The ratio of the masses of two planets is 1 : 10 and the ratio of their diameters is 1 : 2. If the length of a seconds pendulum on the first planet is 0.4 m, then the length of the seconds pendulum on the second planet is _______.

Options

A. 10 cm
B. 0.5 m
C. 10 m
D. 1.0 m

Answer

D

Explanation

`m_(1) : m_(2) = 1 : 10`
`d_(1) : d_(2) = 1 : 2 rArr r_(1) : r_(2) = 1 : 2`
`l_(1) = 0.4 m`
`l_(2) = ?`
`g_(1) = (GM_(1))/(R_(1)^(2)), g_(2) = (GM_(2))/(R_(2)^(2)`
`(g_(1))/(g_(2))=(M_(1))/(R_(1)^(2)) xx (R_(2)^(2))/(M_(2)) = (1)/(10)xx(2^(2))/(1^(2))=(4)/(10)`
`(T_(1))/(T_(2))=sqrt(l_(1)/(g_(1))xx(g_(2))/(l_(2)))=sqrt((0.4)/(l_(2))xx(10)/(4))`
`(2)/(2) = sqrt((1)/(l_(2))),(1)/(l_(2)) = (1)/(1) rArr l_(2) = 1m`